proof of Desargues’ theorem

The claim is that if triangles $A B C$ and $X Y Z$ are perspective from a point $P$ , then they are perspective from a line (meaning that the three points

AB\cdot XY\qquad BC\cdot YZ\qquad CA\cdot ZX

are collinear) and conversely.

Since no three of $A, B, C, P$ are collinear, we can lay down homogeneous coordinates such that

A=(1,0,0)\qquad B=(0,1,0)\qquad C=(0,0,1)\qquad P=(1,1,1)

By hypothesis, there are scalars $p, q, r$ such that

X=(1,p,p)\qquad Y=(q,1,q)\qquad Z=(r,r,1)

The equation for a line through $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ is

(y_{1}z_{2}-z_{1}y_{2})x+(z_{1}x_{2}-x_{1}z_{2})y+(x_{1}y_{2}-y_{1}x_{2})z=0\;,

giving us equations for six lines:

$\displaystyle AB$	$\displaystyle:$	$\displaystyle z=0$
$\displaystyle BC$	$\displaystyle:$	$\displaystyle x=0$
$\displaystyle CA$	$\displaystyle:$	$\displaystyle y=0$
$\displaystyle XY$	$\displaystyle:$	$\displaystyle(pq-p)x+(pq-q)y+(1-pq)z=0$
$\displaystyle YZ$	$\displaystyle:$	$\displaystyle(1-qr)x+(qr-q)y+(qr-r)z=0$
$\displaystyle ZX$	$\displaystyle:$	$\displaystyle(rp-p)x+(1-rp)y+(rp-r)z=0$

whence

$\displaystyle AB\cdot XY$	$\displaystyle=$	$\displaystyle(pq-q,-pq+p,0)$
$\displaystyle BC\cdot YZ$	$\displaystyle=$	$\displaystyle(0,qr-r,-qr+q)$
$\displaystyle CA\cdot ZX$	$\displaystyle=$	$\displaystyle(-rp+r,0,rp-p).$

As claimed, these three points are collinear, since the determinant

\left|\begin{array}[]{ccc}pq-q&-pq+p&0\\ 0&qr-r&-qr+q\\ -rp+r&0&rp-p\end{array}\right|

is zero. (More precisely, all three points are on the line

p(q-1)(r-1)x+(p-1)q(r-1)y+(p-1)(q-1)rz=0\;.)

Since the hypotheses are self-dual, the converse is true also, by the principle of duality.

Title	proof of Desargues’ theorem
Canonical name	ProofOfDesarguesTheorem
Date of creation	2013-03-22 13:47:51
Last modified on	2013-03-22 13:47:51
Owner	drini (3)
Last modified by	drini (3)
Numerical id	5
Author	drini (3)
Entry type	Proof
Classification	msc 51A30